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包括多种数字图像处理变换,有源码。数字图像处理典型算法...
包括多种数字图像处理变换,有源码。数字图像处理典型算法-Including a variety of digital image processing transform, has source code. Typical digital image processing algorithms
- 2022-08-22 16:49:36下载
- 积分:1
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道格拉斯
道格拉斯--普克压缩算法
一种简单的数据压缩算法,采用VB实现.-Douglas Puck compression algorithm a simple data compression algorithm, the use of VB realize.
- 2022-04-30 23:44:39下载
- 积分:1
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用Visual C++实现的小波
利用Visual C++实现的基于wavelet的图像拼接、融合、重建等算法,非常实用-Using Visual C++ implementation of the wavelet-based image mosaic, integration, reconstruction algorithm, a very practical
- 2022-11-28 15:35:03下载
- 积分:1
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模式识别和智能计算的相关软件。注意:这并不是源代码,而是一个软件。但可以了解图像处理的效果。...
模式识别和智能计算的相关软件。注意:这并不是源代码,而是一个软件。但可以了解图像处理的效果。-Pattern recognition and intelligent computing related software. Note: This is not the source code, but a software. But you can understand the image processing results.
- 2022-04-23 01:27:23下载
- 积分:1
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使用opencv对给定图像求直方图。方法简单明了。
使用opencv对给定图像求直方图。方法简单明了。-Opencv use for a given image histogram. Method is simple and clear.
- 2023-02-18 00:35:03下载
- 积分:1
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对图象的对比度进行处理,不同于一般的对比度增强处理
对图象的对比度进行处理,不同于一般的对比度增强处理-the contrast of image processing, from a general contrast enhancement
- 2022-02-01 20:50:14下载
- 积分:1
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图形生成与汉字显示 图形生成与汉字显示
图形生成与汉字显示 图形生成与汉字显示 -GRAPH GENERIZE AND WORD SHOW
- 2022-07-10 05:47:18下载
- 积分:1
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Courseware particle filter, particle filter for a detailed introduction
粒子滤波的课件,对粒子滤波进行了详细的介绍-Courseware particle filter, particle filter for a detailed introduction
- 2022-11-04 17:40:03下载
- 积分:1
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将真彩位图转化成各色或黑白位图,进行二值化(支持多种抖动方法)处理后的位图...
将真彩位图转化成各色或黑白位图,进行二值化(支持多种抖动方法)处理后的位图-of color bitmap into colored or black-and-white bitmap, for the value of two (jitter support multiple methods) treatment, the bitmap
- 2022-10-15 08:25:02下载
- 积分:1
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geometry calculation of correlation algorithm can determine the function of the...
计算几何学的相关算法 可以根据这个函数确定两条线段在交点处的转向,比如确定p0 p1和p1 p2在p1处是左转还是右转,只要求(p2-p0)*(p1-p0),若0则右转,=0则共线-geometry calculation of correlation algorithm can determine the function of the two segments to the intersection, such as identifying p0 p1 p1 and p2 p1 is the left or the right, only requirements (p2- p0)* (p1- p0), and if 0 right, a total = 0 line
- 2022-07-13 05:45:28下载
- 积分:1